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dc.contributor.authorStanley, Richard P.
dc.date.accessioned2015-09-22T11:49:58Z
dc.date.available2015-09-22T11:49:58Z
dc.date.issued2011-02
dc.identifier.issn01956698
dc.identifier.issn1095-9971
dc.identifier.urihttp://hdl.handle.net/1721.1/98849
dc.description.abstractAnswering a question of Bona, it is shown that for n≥2 the probability that 1 and 2 are in the same cycle of a product of two n-cycles on the set {1,2,…,n} is 1/2 if n is odd and 1/2 - 2/(n-1)(n+2) if n is even. Another result concerns the polynomial P[subscript λ](q) = ∑[subscript w]q[superscript κ]((1,2,…,n)⋅w), where w ranges over all permutations in the symmetric group S[subscript n] of cycle type λ, (1,2,…,n) denotes the n-cycle 1→2→⋯→n→1, and κ(v) denotes the number of cycles of the permutation v. A formula is obtained for P[subscript λ](q) from which it is deduced that all zeros of P[subscript λ](q) have real part 0.en_US
dc.description.sponsorshipNational Science Foundation (U.S.) (Grant 0604423)en_US
dc.language.isoen_US
dc.publisherElsevieren_US
dc.relation.isversionofhttp://dx.doi.org/10.1016/j.ejc.2011.01.011en_US
dc.rightsCreative Commons Attribution-Noncommercial-NoDerivativesen_US
dc.rights.urihttp://creativecommons.org/licenses/by-nc-nd/4.0/en_US
dc.sourceMIT Web Domainen_US
dc.titleTwo enumerative results on cycles of permutationsen_US
dc.typeArticleen_US
dc.identifier.citationStanley, Richard P. “Two Enumerative Results on Cycles of Permutations.” European Journal of Combinatorics 32, no. 6 (August 2011): 937–943.en_US
dc.contributor.departmentMassachusetts Institute of Technology. Department of Mathematicsen_US
dc.contributor.mitauthorStanley, Richard P.en_US
dc.relation.journalEuropean Journal of Combinatoricsen_US
dc.eprint.versionAuthor's final manuscripten_US
dc.type.urihttp://purl.org/eprint/type/JournalArticleen_US
eprint.statushttp://purl.org/eprint/status/PeerRevieweden_US
dspace.orderedauthorsStanley, Richard P.en_US
dc.identifier.orcidhttps://orcid.org/0000-0003-3123-8241
mit.licensePUBLISHER_CCen_US
mit.metadata.statusComplete


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